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Question 1 · 7 marks
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(a)
Prove by induction that
∑
r
=
1
n
(
2
r
−
1
)
=
n
2
\sum_{r=1}^{n} (2r - 1) = n^2
∑
r
=
1
n
(
2
r
−
1
)
=
n
2
for all positive integers
n
n
n
.
5 marks
Assume true for
n
=
k
n = k
n
=
k
,
and add the
(
k
+
1
)
(k + 1)
(
k
+
1
)
th term.
∑
r
=
1
k
+
1
(
2
r
−
1
)
=
\sum_{r=1}^{k+1} (2r - 1) =
∑
r
=
1
k
+
1
(
2
r
−
1
)
=
k
xⁿ
a/b
√
π
(
)